Problem: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $13.7$ years; the standard deviation is $3.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living between $20.5$ and $23.9$ years.
Solution: $13.7$ $10.3$ $17.1$ $6.9$ $20.5$ $3.5$ $23.9$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $13.7$ years. We know the standard deviation is $3.4$ years, so one standard deviation below the mean is $10.3$ years and one standard deviation above the mean is $17.1$ years. Two standard deviations below the mean is $6.9$ years and two standard deviations above the mean is $20.5$ years. Three standard deviations below the mean is $3.5$ years and three standard deviations above the mean is $23.9$ years. We are interested in the probability of a lion living between $20.5$ and $23.9$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the lions will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the lions will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of lions between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular lion living between $20.5$ and $23.9$ years is $\color{orange}{2.35\%}$.